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# Polynomial Division

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it is a project to teach people how to do a long division math involving polynomials.

it is a project to teach people how to do a long division math involving polynomials.

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### Transcript

• 1. Dividing the same variable when you divide two terms with a common variable. You subtract the exponent of the division. Ex) x ³ x² = x³-² = x Ex) (2x³ + 4x² + 6x) 2x = 2x³ + 4x² + 6x 2x 2x 2x = 2x² + 2x + 3 Anything to the power of zero is always one
• 2. Simplification and Reduction
• Simplify: x ² + 9x + 14
• x + 7
• This kind a question can be done in either two ways: you can factor the quadratic and then cancel out the common factor, like this.
• x² + 9x + 14
• x + 7
• = (x + 2) (x + 7)
• x + 7
• = x + 2
• You can also solve for this problem by doing long division. Like this
• x + 2
• x + 7 x² + 9x + 14
• - ( x² + 7x )
• = 0 + 2x + 14
• - ( 2x + 14 )
• 0 + 0
• In this case we have no reminder, so the answer is just x+ 2
• 3. Long Division
• 4x ² - 2x +3 quotient
• X + 1 4x ³ + 2x² + x + 5 dividend
• - (4x ³ + 4x² )
• = 0 - 2x ² + x
• - (-2x² - 2x)
• 0 + 3x + 5
• - (3x + 3)
• = 2 this is called the remainder
Divisor
• 4. Dividing polynomial with “missing terms”
• When dividing a polynomial with a missing term, you consider that missing term as a zero. It doesn’t matter if you put a negative sign or positive sign, it is the same.
• ex) 8x ³ + 1
• 2x+ 1
• 4x² - 2x + 1
• 2x + 1 8x³ + 0x² + 0x + 1
• - (8x³ + 4x²)
• = 0 -4x² + 0x
• - ( -4x² - 2x)
• = 0 + 2x + 1
• - (2x + 1)
• 0+ 0
• In this question there is no remainder
• 5. Expression for polynomial “A”
• Ex) A ( x+ 1) – ( 2x + x + 14) = (3x – 2) (x + 4)
• A (x + 1) = (3x – 2) (x+ 4) + (2x + x + 14)
• = (3x ² + 12x -2x – 8)+ (2x + x+ 14)
• A (x + 1) = (3x² + 13x + 6)
• x + 1 (x + 1)
• 3x + 10 R= -4
• x + 1 3x² + 13x + 6
• - (3x² + 3x)
• = 0 + 10x + 6
• - (10x + 10)
• = - 4
• 6. Problem solving
• A B
• D C
• F E
A1= 3x ² - 4x - 7 A2= 4x ² - 3x - 7 x + 1 x + 1 Find DE Find AB
• 7. The diagram above
• A1= 3x² - 4x – 7
• A2= 4x² - 3x – 7
• BC= x + 1
• FE= x + 1
• Find AF, DE, DC, and AB
• To find AF you need to do a long division by using area one and FE
• 3x – 7
• x + 1 3x² - 4x – 7
• - (3x² + 3x)
• = -7x -7
• + ( -7x – 7)
• = o
• so AF= 3x - 7
• 8. The same diagram
• AF= DE + BC
• so since we have AF and BC we can find DE by subtracting AF by BC and that gives us DE
• (3x – 7) – x - 1 = DE
• DE= 2x – 8
• To find DC you need to use area two divide by BC
• 4x – 7
• x + 1 4x² - 3x – 7
• - (4x² + 4x)
• = 0 -7x – 7
• + (-7x – 7)
• = 0
• DC= 4x - 7
• 9. The same diagram
• now we have FE and DC and we are asked to find AB.
• What you do is that you take FE + DC and that gives you AB
• AB = 4x – 7 + x + 1
• = 5x - 6
• 10. Solving for a triangle by using Long Division
• A= 14x ² + x – 3
• B= 4x + 2
• Find the Height
• To solve for this kind a question you need to know that A= B * H
• 2
• So 2A= B* H
• So now you substitute the numbers in
• 2 ( 14x ² + x – 3) = (4x + 2) H
• 28x² + 2x – 6 = (4x + 2) H
• 28x² + 2x – 6 = H
• 4x + 2
• 11. the same triangle
• Now to find the height you need to do long division, like this
• 28x² + 2x – 6
• 4x + 2
• 7x – 3
• 4x + 2 28x² + 2x – 6
• - ( 28x² + 14x)
• = 0 - 12x – 6
• + (-12x – 6)
• = 0
• the height = 7x - 3
• 12. Solving a Rectangle by using long division
• Here are the given information
• A= 8x ² + 22x + 15
• H= 2x + 3
• Find the base of this rectangle
• So to find the base of a rectangle you need to know two things, and those two things are area and the Height.
• So since we have given those two piece of information we can find out what the base is by doing long division.
• 4x + 5
• 2x + 3 8x ² + 22x + 15
• - ( 8x ² + 12x)
• = 10 + 15
• - (10x + 15)
• = 0 + 0